class x chapter 11 arithmetic progression maths

Class X Chapter 11 – Arithmetic Progression Maths

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Exercise – 11A

1. Show that each of the progressions given below is an AP. Find the first term, common

difference and next term of each.

(i) 9, 15, 21, 27,…………

(ii) 11, 6, 1, – 4,……..

(iii) 5 2 1

1, , , ,..........6 3 2

(iv) 2, 8, 18, 32,..........

(v) 20, 45, 80, 125,.........

Sol:

(i) The given progression 9, 15, 21, 27,…………

Clearly, 15 – 9 = 21 – 15 = 27 – 21 = 6 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an

AP.

First term = 9

Common difference =6

Next term of the AP = 27 + 6 = 33

(ii) The given progression 11, 6, 1, – 4,……..

Clearly,6 – 11 = 1 – 6 = –4 – 1 = –5 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an

AP.

First term = 11

Common difference = –5

Next term of the 4 5 9AP

(iii) The given progression5 2 1

1, , , ,..........6 3 2

Clearly, 5 2 5 1 2 1

16 3 6 2 3 6

(Constant)

Thus, each term differs from its preceding term by1

.6

So, the given progression is an

AP.

First term = –1

Common difference = 1

6

Next tern of the 1 1 2 1

2 6 6 3AP


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(iv) The given progression 2, 8, 18, 32,..........

This sequence can be written as 2,2 2,3 2,4 2,........

Clearly, 2 2 2 3 2 2 2 4 2 3 2 2 (Constant)

Thus, each term differs from its preceding term by 2, So, the given progression is

an AP.

First term 2

Common difference 2

Next tern of the 4 2 2 5 2 50AP

(v) This given progression 20, 45, 80, 125,.........

This sequence can be re-written as 2 5,3 5,4 5,5 5,...........

Clearly, 3 5 2 5 4 5 3 5 5 5 4 5 5 (Constant)

Thus, each term differs from its preceding term by 5.So, the given progression is

an AP.

First term 2 5 20

Common difference 5

Next term of the 5 5 5 6 5 180AP

2. Find:

(i) the 20th term of the AP 9,13,17,21,..........

(ii) the 35th term of AP 20,17,14,11,..........

(iii) the 18th term of the AP 2, 18, 50, 98,...........

(iv) the 9th term of the AP 3 5 7 9

, , , ,.........4 4 4 4

(v) the 15the term of the AP 40, 15,10,35,.........

Sol:

(i) The given AP is 9,13,17,21,..........

First term, 9a

Common difference, 13 9 4d thn term of the AP, 1 9 1 4na a n d n

20th term of the AP, 20 9 20 1 4 9 76 85a

(ii) The given AP is 20,17,14,11,..........

First term, 20a

Common difference, 17 20 3d thn term of the AP, 1 20 1 3na a n d n


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35th term of the AP, 35 20 35 1 3 20 102 82a

(iii) The given AP is 2, 18, 50, 98,...........

This can be re-written as 2,3 2,5 2,7 2,.........

First term, 2a

Common difference, 3 2 2 2 2d thn term of the AP, 1 2 1 2 2na a n d n

18th term of the AP, 18 2 18 1 2 2 2 34 2 35 2 2450a

(iv) The given AP is 3 5 7 9

, , , ,.........4 4 4 4

First term, 3

4a

Common difference, 5 3 2 1

4 4 4 2d

thn term of the AP, 3 1

1 14 2

na a n d n

9th term of the AP, 9

3 1 3 199 1 4

4 2 4 4a

(v) The given AP is 40, 15,10,35,.........

First term, 40a

Common difference, 15 40 25d

thn term of the AP, 1 40 1 25na a n d n

15th term of the AP, 15 40 15 1 25 40 350 310a

3. Find the 37th term of the AP3 1 1

6,7 ,9 ,11 ,.........4 2 4

Sol:

The given AP is 3 1 1

6,7 ,9 ,11 ,.........4 2 4

First term, 6a and common difference, 3 31 31 24 7

7 6 64 4 4 4

d

Now, 37 37 1 36T a d a d

76 36 6 63 69

4

37th term = 69


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4. Find the 25th term of the AP1 1

5, 4 , 4,3 ,3,........2 2

Sol:

The given AP is 1 1

5, 4 , 4,3 ,3,........2 2

First term = 5

Common difference 1 9 9 10 1

4 5 52 2 2 2

15

2a and d

Now, 25 25 1 24T a d a d

15 24 5 12 7

2

25th term 7

5. Find the nth term of each of the following Aps:

(i) 5, 11, 17, 23 …

(ii) 16, 9, 2, -5, …..

Sol:

(i) (6n – 1)

(ii) (23 – 7n)

6. If the nth term of a progression is (4n – 10) show that it is an AP. Find its

(i) first term, (ii) common difference (iii) 16 the term.

Sol:

4 10n nT [Given]

1 4 1 10 6T

2

3

4

4 2 10 2

4 3 10 2

4 4 10 6

T

T

T

Clearly, 2 6 2 2 6 2 4 (Constant)

So, the terms 6, 2,2,6,...... forms an AP.

Thus we have

(i) First term 6

(ii) Common difference = 4

(iii) 16 1 15 6 15 4 54T a n d a d


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7. How many terms are there in the AP 6,1 0, 14, 18, ….., 174?

Sol:

In the given AP, 6a and 10 6 4d

Suppose that there are n terms in the given AP.

Then, 174nT

1 174a n d

6 1 4 174

2 4 174

4 172

43

n

n

n

n

Hence, there are 43 terms in the given AP.

8. How many terms are there in the AP 41, 38, 35, ….,8?

Sol:

In the given AP, 41 38 41 3a and d

Suppose that there are n terms in the given AP.

Then 8nT

1 8

41 1 3 8

44 3 8

3 36

12

a n d

n

n

n

n


9. How many terms are there in the AP 1

18,15 ,13,....., 47.2

?

Sol:

The given AP is 1

18,15 ,13,....., 47.2

First term, 18a

Common difference, 1 31 31 36 5

15 18 182 2 2 2

d

Suppose there re n terms in the given AP. Then,


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47

518 1 47 1

2

51 47 18 65

2

21 65 26

5

26 1 27

n

n

a

n a a n d

n

n

n


10. Which term of term of the AP 3,8,13,18,….. is 88?

Sol:

In the given AP, first term, 3a and common difference, 8 3 5.a

Let’s its thn term be 88

Then, 88nT

1 88

3 1 5 88

5 2 88

5 90

18

a n d

n

n

n

n

Hence, the 18thterm of the given AP is 88.

11. Which term of AP 72,68,64,60,… is 0?

Sol:

In the given AP, first term, 72a and common difference, 68 72 4.d

Let its thn term be 0.

Then, 0nT

1 0

72 1 4 0

76 4 0

4 76

19

a n d

n

n

n

n

Hence, the 19thterm of the given AP is 0.


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12. Which term of the AP 5 1

,1,1 ,1 ,.....6 6 3

is 3?

Sol:

In the given AP, first term5

6 and common difference,

5 11

6 6d

Let its thn term be 3.

Now, 3nT

1 3

5 11 3

6 6

23

3 6

7

6 3

14

a n d

n

n

n

n

Hence, the 14th term of the given AP is 3.

13. Which term of the AP 21, 18, 15, …… is -81?

Sol:

The given AP is 21,18,15,.......

First term, 21a


Suppose thn term of the given AP is 81. then,

81

21 1 3 81 1

3 1 81 21 102

1021 34

3

34 1 35

n

n

a

n a a n d

n

n

n

Hence, the 35th term of the given AP is -81.

14. Which term of the AP 3,8, 13,18,…. Will be 55 more than its 20th term?

Sol:

Here, 3a and 8 3 5d

The 20th term is given by

20 20 1 19 3 19 5 98T a d a d

Required term 98 55 153


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Let this be the thn term.

Then 153nT

3 1 5 153

5 155

31

n

n

n

Hence, the 31st term will be 55 more than 20th term.

15. Which term of the AP 5, 15, 25, ….. will be 130 more than its 31st term?

Sol:

Here, 5a and 15 5 10d

The 31st term is given by

31 31 1 30 5 30 10 305T a d a d

Required term 305 130 435

Let this be the thn term.

Then, 435nT

5 1 10 435

10 440

44

n

n

n

Hence, the 44th term will be 130 more than its 31st term.

16. If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP

Sol:

In the given AP, let the first term be a and the common difference be d.

Then, 1nT a n d

Now, we have:

10 10 1T a d

9 52 ........ 1a d

13

17

13 1 12 .... 2

17 1 16 .... 3

T a d a d

T a d a d

But, it is given that 17 1320T T

i.e., 16 20 12a d a d

4 20

5

d

d

On substituting 5d in (1), we get:


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9 5 52

7

a

a

Thus, 7 5a and d

The terms of the AP are 7,12,17,22,.........

17. Find the middle term of the AP 6, 13, 20, …., 216.

Sol:

The given AP is 6,13,20,........,216.

First term, 6a


Suppose these are n terms in the given AP. Then,

216

6 1 7 216 1

7 1 216 6 210

2101 30

7

n

n

a

n a a n d

n

n

30 1 31n

Thus, the given AP contains 31 terms,

Middle term of the given AP

31 1

2th

term

16th term

6 16 1 7

6 105

111

Hence, the middle term of the given AP is 111.

18. Find the middle term of the AP 10, 7, 4, ……., (-62).

Sol:

The given AP is 10,7,4,....., 62.

First term, 10a


Suppose these are n terms in the given AP. Then,


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62

10 1 3 62 1

3 1 62 10 72

721 24

3

24 1 25

n

n

a

n a a n d

n

n

n

Thus, the given AP contains 25 terms.

Middle term of the given AP

25 1

2th

term

13th term

10 13 1 3

10 36

26

Hence, the middle term of the given AP is 26.

19. Find the sum of two middle most terms of the AP 4 2 1

, 1, ,..., 4 .3 3 3

Sol:


, 1, ,....., 4 .3 3 3

First term,4

3a

Common difference,4 4 1

1 13 3 3

d

Suppose there are n terms in the given AP. Then,

14

3

4 1 131 1

3 3 3

1 13 4 171

3 3 3 3

1 17

17 1 18

n

n

a

n a a n d

n

n

n

Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.

The middle terms of the given AP are 18

2th

terms and 18

12

th

term i.e. 9th term and

10th term.


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Sum of the middle most terms of the given AP

= 9th term + 10th term

4 1 4 1

9 1 10 13 3 3 3

4 8 43

3 3 3

3

Hence, the sum of the middle most terms of the given AP is 3.

20. Find the 8th term from the end of the AP 7, 10, 13, ……, 184.

Sol:

Here, 7a and 10 7 3, 184d l and 8thn form the end.

Now, nth term from the end 1l n d

8th term from the end 184 8 1 3

184 7 3 184 21 163

Hence, the 8th term from the end is 163.

21. Find the 6th term form the end of the AP 17, 14, 11, ……, (-40).

Sol:

Here, 7a and 14 17 3, 40 6d l and n

Now, nth term from the end 1l n d

6th term from the end 40 6 1 3

40 5 3 40 15 25

Hence, the 6th term from the end is –25.

22. Is 184 a term of the AP 3, 7, 11, 15, ….?

Sol:

The given AP is 3,7,11,15,......

Here, 3a and 7 3 4d

Let the nth term of the given AP be 184. Then,

184na

3 1 4 184 1

4 1 184

nn a a n d

n


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4 185

185 146

4 4

n

n

But, the number of terms cannot be a fraction.

Hence, 184 is not a term of the given AP.

23. Is -150 a term of the AP 11, 8, 5, 2, ……?

Sol:

The given AP is 11, 8, 5, 2, ……


Let the nth term of the given AP be 150. Then,

150

11 1 3 150 1

3 14 150

3 164

164 254

3 3

n

n

a

n a a n d

n

n

n

But, the number of terms cannot be a fraction.

Hence, -150 is not a term of the given AP.

24. Which term of the AP 121, 117, 113, …. is its first negative term?

Sol:

The given AP is 121, 117, 113, …….

Here, 121a and 117 121 4d

Let the nth term of the given AP be the first negative term. Then,

0

121 1 4 0 1

n

n

a

n a a n d

125 4 0

4 125

n

n

125 131

4 4

32

n

n

Hence, the 32nd term is the first negative term of the given AP.


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25. Which term of the AP 1 1 3

20,19 ,18 ,17 ,.....4 2 4

is the first negative term?

Sol:


20,19 ,18 ,17 ,.....4 2 4

Here, 20a and 1 77 77 80 3

19 20 204 4 4 4

d


0

320 1 0 1

4

3 320 0

4 4

83 30

4 4

3 83

4 4

83 227

3 3

28

n

n

a

n a a n d

n

n

n

n

n

Hence, the 28th term is the first negative term of the given AP.

26. The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.

Sol:

We have

7 1

6 4 ........ 1

T a n d

a d

13 1

12 16 ....... 2

T a n d

a d

On solving (1) and (2), we get

8 2a and d

Thus, first term = 8 and common difference 2

The term of the AP are 8,6,4,2,.........

27. The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.

Sol:

In the given AP, let the first be a and the common difference be d.

Then, 1nT a n d


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Now, 4 4 1T a d

3 0 ...... 1

3

a d

a d

Again, 11 11 1 10T a d a d

3 10 7d d d [Using (1)]

Also, 25 25 1 24 3 24 21T a d a d d d d [Using (1)]

i.e., 25 113 7 3T d T

Hence, 25th term is triple its 11th term.

28. The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.

Sol:

Let a be the first term and d be the common difference of the AP. Then,

8 0 1

8 1 0

7 0

7 ....... 1

na a a n d

a d

a d

a d

Now,

38

18

38

18

38

18

38 18

38 1

18 1

7 37(1)

7 17

303

10

3

a da

a a d

a d dFrom

a d d

a d

a d

a a

Hence, the 38th term of the AP id triple its 18th term.

29. The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its

common difference

Sol:


4 11

4 1 11 1

3 11 ........ 1

n

a

a d a a n d

a d

Now,

5 7 34a a (Given)


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4 6 34

2 10 34

5 17 ...... 2

a d a d

a d

a d

From (1) and (2), we get

11 3 5 17

2 17 11 6

3

d d

d

d

Hence, the common difference of the AP is 3.

30. The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common

difference of the AP.

Sol:


9 32

9 1 32 1

8 32 ........ 1

n

a

a d a a n d

a d

Now,

11 13 94a a (Given)

10 12 94

2 22 94

11 47 ....... 2

a d a d

a d

a d


32 8 11 47

3 47 32 15

5

d d

d

d


31. Determine the nth term of the AP whose 7th term is -1 and 16th term is 17.

Sol:


7 1a

7 1 1 1

6 1 .......... 1

na d a a n d

a d

Also,

16 17

15 17 ........ 2

a

a d


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1 6 15 17

9 17 1 18

2

d d

d

d

Putting 2d in (1), we get

6 2 1

1 12 13

a

a

1

13 1 2

2 15

na a n d

n

n

Hence, the nth term of the AP is 2 15 .n

32. If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.

Sol:


4 184 18a a (Given)

4 3 18 17 1

2 3 9 17

2 6 9 153

7 147

21

21 0

22 1 0

na d a d a a n d

a d a d

a d a d

a d

a d

a d

a d

22 0a

Hence, the 22nd term of the AP is 0.

33. If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term

is zero.

Sol:


10 1510 15a a (Given)

10 9 15 14 1na d a d a a n d

2 9 3 14

2 18 3 42

24

a d a d

a d a d

a d


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25

24 0

25 1 0

0

a d

a d

a

Hence, the 25th term of the AP is 0.

34. Find the common difference of an AP whose first term is 5 and the sum of its first four

terms is half the sum of the next four terms.

Sol:

Let the common difference of the AP be d.

First term, 5a

Now,

1 2 3 4 5 6 7 8

1

2a a a a a a a a (Given)

1

2 3 4 5 6 72

a a d a d a d a d a d a d a d

1na a n d

1

4 6 4 222

a d a d

8 12 4 22

22 12 8 4

10 4

2

5

25 2 5

5

a d a d

d d a a

d a

d a

d a


35. The sum of the 2nd and 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th

term, find AP.

Sol:


2 7 30a a (Given)

6 30 1

2 7 30 .............. 1

na d a d a a n d

a d

Also,

15 82 1a a (Given)


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14 2 7 1

14 2 14 1

1

1

a d a d

a d a d

a

a

Putting a = 1in (1), we get

2 1 7 30

7 30 2 28

4

d

d

d

So,

2

3

1 4 5

2 1 2 4 9..........

a a d

a a d

Hence, the AP is 1, 5, 9, 13, …….

36. For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, … and 3, 10,

17, … are equal?

Sol:

Let the term of the given progressions be nt and ,nT respectively.

The first AP is 63, 65, 67,...

Let its first term be a and common difference be d.

Then a = 63 and d = (65 - 63) = 2

So, its nth term is given by

1

63 1 2

61 2

nt a n d

n

n

The second AP is 3, 10, 17,...

Let its first term be A and common difference be D.

Then A = 3 and D = (10 - 3) = 7

So, its nth term is given by

1

3 1 7

7 4

nT A n D

n

n

Now, n nt T

61 2 7 4

65 5

13

n n

n

n

Hence, the l3 terms of the Al’s are the same.


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37. The 17th term of AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find

its nth term.

Sol:


17 82 5a a (Given)

16 2 7 5 1

16 2 14 5

2 5 ......... 1

na d a d a a n d

a d a d

a d

Also,

11 43a (Given)

10 43 ........ 2a d


5 2 10 43

12 43 5 48

4

d d

d

d


2 4 5

5 8 3

1

3 1 4

4 1

n

a

a

a a n d

n

n

Hence, the nth term of the AP is 4 1 .n

38. The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Sol:


24 102a a (Given)

23 2 9 1

23 2 18

2 23 18

5 ....... 1

na d a d a a n d

a d a d

a a d d

a d

Now,

72

15

71

14

a a d

a a d


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72

15

5 71(1)

5 14

a d dFrom

a d d

72

15

72 15

764

19

4

a d

a d

a a

Hence, the 72nd term of the AP is 4 times its 15th term.

39. The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.

Sol:


19 63a a (Given)

18 3 5 1

18 3 15

3 18 15

2 3 ........... 1

na d a d a a n d

a d a d

a a d d

a d

Also,

9 19a (Given)

8 19 ...... 2a d


38 19

2

3 1619

2

19 38

2

dd

d d

d

d


2 3 2 6

3

a

a

So,

2

3

3 2 5

2 3 2 2 7,........

a a d

a a d

Hence, the AP is 3,5,7,9,........

40. If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.

Sol:

In the given AP, let the first be a and the common difference be d.


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Then, 1nT a n d

1 .......Tp a p d q i

1 ........qT a q d p ii

On subtracting (i) from (ii), we get:

1

q p d p q

d

Putting 1d in (i), we get:

1a p q

Thus, 1 1a p q and d

Now, 1p qT a p q d

1 1 1

1 1 0

p q p q

p q p q

Hence, the th

p q term is 0 (zero).

41. The first and last terms of an AP are a and l respectively. Show that the sum of the nth term

from the beginning and the nth term form the end is ( a + l ).

Sol:

In the given AP, first term = a and last term = l.

Let the common difference be d.

Then, nth term from the beginning is given by

1 ..... 1nT a n d

Similarly, nth term from the end is given by

1 ...... 2nT l n d

Adding (1) and (2), we get

1 1

1 1

1

a n d l n d

a n d l n d

a

Hence, the sum of the nth term from the beginning and the nth term from the end 1 .a

42. How many two-digit number are divisible by 6?

Sol:

The two digit numbers divisible by 6 are 12, 18, 24,……, 96

Clearly, these number are in AP.


Let this AP contains n terms. Then,


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96

12 1 6 96 1

6 6 96

6 96 6 90

15

n

n

a

n a a n d

n

n

n

Hence, these are 15 two-digit numbers divisible by 6.

43. How many two-digits numbers are divisible by 3?

Sol:

The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.




99

12 1 3 99 1

3 9 99

3 99 9 90

30

n

n

a

n a a n d

n

n

n

Hence, there are 30 two-digit numbers divisible by 3.

44. How many three-digit numbers are divisible by 9?

Sol:

The three-digit numbers divisible by 9 are 108, 117, 126,...., 999.


Here. a = 108 and d = 117 – 108 = 9

Let this AP contains n terms. Then.

999

108 1 9 999 1

9 99 999

9 999 99 900

100

n

n

a

n a a n d

n

n

n

Hence: there are 100 three-digit numbers divisible by 9.

45. Hoe many numbers are there between 101 and 999, which are divisible by both 2 and 5?

Sol:

The numbers which are divisible by both 2 and 5 are divisible by 10 also.

Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, .., 990.

Clearly, these number are in AP


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Here, 110a and 120 110 10d


990

110 1 10 990 1

10 100 990

10 990 100 890

89

n

n

a

n a a n d

n

n

n

Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

46. In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and

so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?

Sol:

The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11.

Difference of rose plants between two consecutive rows = (41 – 43) = (39 – 41) = –2

[Constant]

So, the given progression is an AP

Here, first term = 43

Common difference = –2

Last term 11

Let n be the last term, then we have:

1

11 43 1 2

11 45 2

34 2

17

nT a n d

n

n

n

n

Hence, the 17th term is 11 or there are 17 rows in the flower bed.

47. A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less

than the preceding prize, find the value of each of the prizes.

Sol:

Let the amount of the first prize be ₹a

Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the

four prizes are in AP.

Amount of the second prize ₹ 200a

Amount of the third prize ₹ 2 200 400a a

Amount of the fourth prize ₹ 3 200 600a a

Now,


______________________________________________________________________________

Total sum of the four prizes = 2,800

₹a+₹ 200a ₹ 400a ₹ 600a ₹2,800

4 1200 2800

4 2800 1200 4000

1000

a

a

a

Amount of the first prize = ₹1,000

Amount of the second prize ₹ 1000 200 =₹800

Amount of the third prize ₹ 1000 400 ₹600

Amount of the fourth prize ₹ 1000 600 = ₹400

Hence, the value of each of the prizes is ₹1,000, ₹800, ₹600 and ₹400.

Exercise - 11B

1. Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.

Sol:

It is given that 3 2 , 4 6k k and 2k are three consecutive terms of an AP.

4 6 3 2 2 4 6

4 6 3 2 2 4 6

4 3 8

3 8 4

4 12

3

k k k k

k k k k

k k

k k

k

k

Hence, the value of k is 3.

2. Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.

Sol:

It is given that 5 2 , 4 1x x and 2x are in AP.

4 1 5 2 2 4 1

4 1 5 2 2 4 1

3 3 3

3 3 3

2 6

3

x x x x

x x x x

x x

x x

x

x

Hence, the value of x is 3.


______________________________________________________________________________

3. If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value

of y.

Sol:

It is given that 3 1 , 3 5y y and 5 1y are three consecutive terms of an AP.

3 5 3 1 5 1 3 5

3 5 3 1 5 1 3 5

6 2 4

2 6 4 10

5

y y y y

y y y y

y

y

y

Hence, the value of y is 5.

4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.

Sol:

Since 2 ,2x x and 2 3x are in AP, we have

2 2 2 3 2

2 3

5

5

x x x x

x

x

x

5. Show that 2 2 2,a b a b and 2 2a b are in AP.

Sol:

The given numbers are 2 2 2,a b a b and

2.a b

Now,

22 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2

2 2

a b a b a b a ab b a b a ab b ab

a b a b a ab b a b ab

So, 2 22 2 2 2 2a b a b a b a b ab (Constant)

Since each term differs from its preceding term by a constant, therefore, the given numbers

are in AP.

6. Find the three numbers in AP whose sum is 15 and product is 80.

Sol:

Let the required numbers be ,a d a and .a d

Then 15a d a a d


______________________________________________________________________________

3 15

5

a

a

Also, . . 80a d a a d

2 2

2

2

80

5 25 80

25 16 9

3

a a d

d

d

d

Thus, 5a and 3d

Hence, the required numbers are 2,5 8 8,5 2 .and or and

7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers.

Sol:

Let the required numbers be ,a d a and .a d

Then 3a d a a d

3 3

1

a

a


2 2

2

2

35

1. 1 35

36

6

a a d

d

d

d

Thus, 1a and 6d

Hence, the required numbers are 5,1 7 7,1 5 .and or and

8. Divide 24 in three parts such that they are in AP and their product is 440.

Sol:

Let the required parts of 24 be ,a d a and a d such that they are in AP.

Then 24a d a a d

3 24

8

a

a


2 2

2

440

8 64 440

a a d

d


______________________________________________________________________________

2 64 55 9

3

d

d

Thus, 8a and 3d

Hence, the required parts of 24 are 5,8,11 11,8,5 .or

9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these

terms is 165. Find these terms

Sol:

Let the required terms be ,a d a and .a d

Then 21a d a a d

3 21

7

a

a

Also, 2 22 165a d a a d

2 2

2

2

2

3 2 165

3 49 2 165

2 165 147 18

9

3

a d

d

d

d

d

Thus, 7a and 3d

Hence, the required terms are 4,7,10 10,7,4 .or

10. The angles of quadrilateral are in whose AP common difference is 10 . Find the angles.

Sol:

Let the required angles be 15 , 5 , 5 15 ,a a a and a as the common

difference is 10 (given).

Then 15 5 5 15 360a a a a

4 360

90

a

a

Hence, the required angles of a quadrilateral are

90 15 , 90 5 , 90 5 90 15 ; 75 ,85 ,95 105 .and or and

11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.

Sol:

(4, 6, 8, 10) or (10, 8, 6, 4)


______________________________________________________________________________

12. Divide 32 into four parts which are the four terms of an AP such that the product of the first

and fourth terms is to product of the second and the third terms as 7:15.

Sol:

Let the four parts in AP be 3 , , 3 .a d a d a d and a d Then,

3 3 32

4 32

8 ......... 1

a d a d a d a d

a

a

Also,

2

2

2 2

2 2

2

2

2

3 3 : 7 :15

8 3 8 3 7(1)

8 8 15

64 9 7

64 15

15 64 9 7 64

960 135 448 7

135 7 960 448

128 512

4

2

a d a d a d a d

d dFrom

d d

d

d

d d

d d

d d

d

d

d

When 8a and 2,d

3 8 3 2 8 6 2

8 2 6

8 2 10

3 8 3 2 8 6 14

a d

a d

a d

a d

When 8a and 2,d

3 8 3 2 8 6 14

8 2 8 2 10

8 2 6

3 8 3 2 8 6 2

a d

a d

a d

a d

Hence, the four parts are 2, 6, 10 and 14.

13. The sum of first three terms of an AP is 48. If the product of first and second terms exceeds

4 times the third term by 12. Find the AP.

Sol:

Let the first three terms of the AP be , .a d a and a d Then,


______________________________________________________________________________

48

3 48

16

a d a a d

a

a

Now,

4 12a d a a d (Given)

16 16 4 16 12

256 16 64 4 12

16 4 256 76

20 180

9

d d

d d

d d

d

d

When 16a and 9d ,

16 9 7

16 9 25

a d

a d

Hence, the first three terms of the AP are 7, 16, and 25.

Exercise - 11C

1. The first three terms of an AP are respectively (3y – 1), (3y + 5) and (5y + 1), find the value

of y.

Sol:

The terms 3 1 , 3 5y y and 5 1y are in AP.

3 5 3 1 5 1 3 5

3 5 3 1 5 1 3 5

6 2 4

2 10

5

y y y y

y y y y

y

y

y

Hence, the value of y is 5.

2. If , 2 1k k and 2 1k are the three successive terms of an AP, find the value of k.

Sol:

It is given that , 2 1k k and 2 1k are the three successive terms of an AP.

2 1 2 1 2 1

1 2

3

k k k k

k

k

Hence, the value of k is 3.


______________________________________________________________________________

3. If 18, a, (b - 3) are in AP, then find the value of (2a – b)

Sol:

It is given that 18, , 3a b are in AP.

18 3

18 3

2 15

a b a

a a b

a b

Hence, the required value is 15.

4. If the numbers a, 9, b, 25 from an AP, find a and b.

Sol:

It is given that the numbers ,9, ,25a b from an AP.

9 9 25a b b

So,

9 25

2 34

17

b b

b

b

Also,

9 9

18

18 17 17

1

a b

a b

a b

a

Hence, the required values of a and b are 1 and 17, respectively.

5. If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers

Sol:

It is given that the numbers 2 1 , 3 2n n and 6 1n are in AP.

3 2 2 1 6 1 3 2

3 2 2 1 6 1 3 2

3 3 3

2 6

3

n n n n

n n n n

n n

n

n

When, 3n

2 1 2 3 1 6 1 5

3 2 3 3 2 9 2 11

6 1 6 3 1 18 1 17

n

n

n

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.


______________________________________________________________________________

6. How many three-digit natural numbers are divisible by 7?

Sol:

The three digit natural numbers divisible by 7 are 105,112,119......,994


Here, 105 112 105 7a and d


994

105 1 7 994 1

7 98 994

7 994 98 986

128

n

n

a

n a a n d

n

n

n

Hence, there are 128 three digit numbers divisible by 7.

7. How many three-digit natural numbers are divisible by 9?

Sol:

The three-digit natural numbers divisible by 9 are 108, 117, 126 ,….. 999.


Here. a = 108 and d = 117 – 108 = 9


999

108 1 9 999 1

9 99 999

9 999 99 900

100

n

n

a

n a a n d

n

n

n

Hence, there are 100 three-digit numbers divisible by 9.

8. If the sum of first m terms of an AP is ( 22 3m m ) then what is its second term?

Sol:

Let mS denotes the sum of first m terms of the AP.

2

2 2 2

1

2 3

2 1 3 1 2 2 1 3 1 2 1

m

m

S m m

S m m m m m m m

Now, thm term of AP, 1m m ma S S

2 22 3 2 1 4 1ma m m m m m

Putting 2,m we get

2 4 2 1 9a


______________________________________________________________________________

Hence, the second term of the AP is 9.

9. What is the sum of first n terms of the AP a, 3a, 5a, …..

Sol:

The given AP is 3 ,5 ,........a a

Here,

First term, A = a

Common difference, 3 2D a a a

Sum of the n terms, nS

2

2 1 2 2 12 2

2 2 22

22

n

n na n a S A n D

na an a

nan

an

Hence, the required sum is 2.an

10. What is the 5th term form the end of the AP 2, 7, 12, …., 47?

Sol:

The given AP is 2, 7, 12, ..., 47.

Let us re-write the given AP in reverse order i.e. 47, 42, .., 12, 7, 2.

Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of

the AP 47, 42,.... ,12, 7, 2.

Consider the AP 47, 42,..., 12, 7, 2.

Here, a = 47 and d = 42 – 47 = –5

5th term of this AP

= 47 + (5 – 1) (–5)

= 47 – 20

= 27

Hence, the 5th term from the end of the given AP is 27.

11. If na denotes the nth term of the AP 2, 7, 12, 17, … find the value of 30 20a a .

Sol:

The given AP is 2, 7, 12, 17,………

Here, 2a and 7 2 5d


______________________________________________________________________________

30 20

2 30 1 5 2 20 1 5 1

147 97

50

n

a a

a a n d

Hence, the required value is 50.

12. The nth term of an AP is (3n +5 ). Find its common difference.

Sol:

We have

3 5nT n

Common difference 2 1T T

1

2

3 1 5 8

3 2 5 11

T

T

11 8 3d

Hence, the common difference is 3.

13. The nth term of an AP is (7 – 4n). Find its common difference.

Sol:

We have

7 4nT n

Common difference 2 1T T

1

2

7 4 1 3

7 4 2 1

1 3 4

T

T

d

Hence, the common difference is .4

14. Write the next term for the AP 8, 18, 32,.........

Sol:

The given AP is 8, 18, 32,.........

On simplifying the terms, we get:

2 2,3 2,4 2,......

Here, 2 2a and 3 2 2 2 2d

Next term, 4 3 2 2 3 2 5 2 50T a d


______________________________________________________________________________

15. Write the next term of the AP 2, 8, 18,.........

Sol:

The given AP is 2, 8, 18,.........

On simplifying the terms, we get:

2,2 2,3 2,......

Here, 2a and 2 2 2 2d

Next term, 4 3 2 3 2 4 2 32T a d

16. Which term of the AP 21, 18, 15, … is zero?

Sol:

In the given AP, first term, 21a and common difference, 18 21 3d

Let’s its thn term be 0.

Then, 0nT

1 0a n d

21 1 3 0

24 3 0

3 24

8

n

n

n

n

Hence, the 8th term of the given AP is 0.

17. Find the sum of the first n natural numbers.

Sol:

The first n natural numbers are 1, 2, 3, 4, 5,……..,n

Here, a = 1 and d = (2 – 1) = 1

Sum of n terms of an AP is given by

2 12

2 1 1 12

12 1 1

2 2 2

n

nS a n d

nn

n nn nn n

18. Find the sum of first n even natural numbers.

Sol:

The first n even natural numbers are 2, 4, 6, 8, 10, …., n.


______________________________________________________________________________

Here, a = 2 and 4 2 2d

Sum of n terms of an AP is given by

2 12

2 2 1 22

4 2 2 2 2 12 2

n

nS a n d

nn

n nn n n n

Hence, the required sum is 1 .n n

19. The first term of an AP is p and its common difference is q. Find its 10th term.

Sol:

Here, a p and d q

Now, 1nT a n d

10

1

9

nT p n q

T p q

20. If 4

, , 25

a are in AP, find the value of a.

Sol:

If 45, a and 2 are three consecutive terms of an AP, then we have:

45 2

2 2 45

2 145

75

a a

a

a

a

21. If (2p +1), 13, (5p -3) are in AP, find the value of p.

Sol:

Let 2 1 ,13, 5 3p p be three consecutive terms of an AP.

Then 13 2 1 5 3 13p p

7 28

4

p

p

When 4, 2 1 ,13p p and 5 3p from three consecutive terms of an AP.


______________________________________________________________________________

22. If (2p – 1), 7, 3p are in AP, find the value of p.

Sol:

Let 2 1 ,7p and 3p be three consecutive terms of an AP.

Then 7 2 1 3 7p p

5 15

3

p

p

When 3, 2 1 ,7p p and 3p form three consecutive terms of an AP.

23. If the sum of first p terms of an AP is 2( ),ap bp find its common difference.

Sol:

Let pS denotes the sum of first p terms of the AP.

2

2

1

2

2

1 1

2 1 1

2

p

p

S ap bp

S a p b p

a p p b p

ap a b p a b

Now, thp term of 1, p p pAP a S S

2 2

2 2

2

2

2

ap bp ap a b p a b

ap bp ap a b p a b

ap a b

Let d be the common difference of the AP.

1

2 2 1

2 2 1

2

p pd a a

ap a b a p a b

ap a b a p a b

a

Hence, the common difference of the AP is 2a.

24. If the sum of first n terms is 2(3 5 ),n n find its common difference.

Sol:

Let nS denotes the sum of first n terms of the AP.

2

2

1

3 5

3 1 5 1

n

n

S n n

S n n


______________________________________________________________________________

23 2 1 5 1n n n

23 2n n

Now, thn term of 1, n n nAP a S S

2 23 5 3 2

6 2

n n n n

n


1

6 2 6 1 2

6 2 6 1 2

6

n nd a a

n n

n n


25. Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.

Sol:


4 9

4 1 9 1

3 9 ......... 1

n

a

a d a a n d

a d

Now,

6 13 40a a (Given)

5 12 40a d a d

2 17 40 ......... 2a d


2 9 3 17 40

18 6 17 40

11 40 18 22

2

d d

d d

d

d


3 2 9

9 6 3

a

a

Hence, the AP is 3, 5, 7, 9, 11,…….


______________________________________________________________________________

Exercise – 11D

1. Find the sum of each of the following Aps:

(i) 2, 7, 12, 17, ……. to 19 terms.

(ii) 9, 7, 5, 3 … to 14 terms

(iii) -37, -33, -29, … to 12 terms.

(iv) 1 1 1

, , ,.....15 12 10

to 11 terms.

(v) 0.6, 1.7, 2.8, …. to 100 terms

Sol:

(i) The given AP is 2, 7, 12, 17,………

Here, 2a and 7 2 5d

Using the formula, 2 1 ,2

n

nS a n d we have

19

192 2 19 1 5

2

194 90

2

1994

2

893

S

(ii) The given AP is 9, 7, 5, 3,……….

Here, 9a and 7 9 2d


n

nS a n d we have

14

142 9 14 1 2

2

7 18 26

7 8

56

S

(iii) The given AP is -37, -33, -29,……..

Here, 37a and 33 37 33 37 4d


n

nS a n d we have

12

122 37 12 1 4

2S

6 74 44

6 30


______________________________________________________________________________

180

(iv) The given AP is 1 1 1

, , ,......15 12 10

Here, 1

15a and

1 1 5 4 1

12 15 60 60d


n

nS a n d we have

11

11 1 12 11 1

2 15 60

11 2 10

2 15 60

S

11 18

2 60

33

20

(v) The given AP is 0.6,1.7,2.8,..........

Here, 0.6a and 1.7 0.6 1.1d

Using formula, 2 1 ,2

n

nS a n d we have

100

1002 0.6 100 1 1.1

2S

50 1.2 108.9

50 110.1

5505

2. Find the sum of each of the following arithmetic series:

(i) 1

7 10 14 ... 842

(ii) 34 32 30 ... 10

(iii) ( 5) ( 8) ( 11) ... ( 230)

Sol:

(i) The given arithmetic series is 1

7 10 14 ..... 84.2

Here, 1 21 21 4 7

7, 10 7 72 2 2 2

a d

and 84.l

Let the given series contains n terms. Then,


______________________________________________________________________________

84

77 1 84 1

2

7 784

2 2

7 7 16184

2 2 2

16123

7

n

n

a

n a a n d

n

n

n

Required sum 23

7 842 2

n

nS a l

2391

2

2030

2

11046

2

(ii) The given arithmetic series is 34 32 30 ..... 10.

Here, 34, 32 34 2a d and 10.l

Let the given series contain n terms. Then,

10

34 1 2 10 1

2 36 10

2 10 36 26

13

n

n

a

n a a n d

n

n

n

Required sum 13

34 102 2

n

nS a l

1344

2

286

(iii) The given arithmetic series is 5 8 11 ....... 230 .

Here, 5, 8 5 8 5 3a d and 230.l

Let the given series contain n terms. Then,


______________________________________________________________________________

230

5 1 3 230 1

3 2 230

3 230 2 228

76

n

n

a

n a a n d

n

n

n

Required sum 76

5 2302 2

n

nS a l

76

2352

8930

3. Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its

first 20 terms.

Sol:

Let na be the nth term of the AP.

5 6na n

Putting 1,n we get

First term, 1 5 6 1 1a a

Putting 2,n we get

2 5 6 2 7a


2 1 7 ( 1) 7 1 6d a a

Sum of first n tern of the AP, nS

2

2 1 1 6 2 12 2

2 6 62

2 3

2 3

n

n nn S a n d

nn

n n

n n

Putting 20,n we get

2

20 2 20 3 20 40 1200 1160S

4. The sum of the first n terms of an AP is 23 6 .n n Find the nth term and the 15th term of this

AP.

Sol:


______________________________________________________________________________


2

2

1

2

2

3 6

3 1 6 1

3 2 1 6 1

3 3

n

n

S n n

S n n

n n n

n

thn term of the AP, na

1

2 23 6 3 3

6 3

n nS S

n n n

n

Putting 15,n we get

15 6 15 3 90 3 93a

Hence, the thn term is 6 3n and 15th term is 93.

5. The sum of the first n terms of an AP is given by 23nS n n . Find its

(i) nth term,

(ii) first term and

(iii) common difference.

Sol:

Given: 23 ......nS n n i

Replacing n by 1n in (i), we get:

2

1

2

2

3 1 1

3 2 1 1

3 7 4

nS n n

n n n

n n

(i) Now, 1n n nT S S

2 23 3 7 4 6 4n n n n n

thn term, 6 4 .......nT n ii

(ii) Putting 1n in (ii), we get:

1 6 1 4 2T

(iii) Putting 2n in (ii), we get:

2 6 2 4 8T

Common difference, 2 1 8 2 6d T T


______________________________________________________________________________

6. The sum of the first n terms of an AP is 25 3

2 2

n n

. Find its nth term and the 20th term of

this AP.

Sol:

2

25 3 15 3 .......

2 2 2n

n nS n n i

Replacing n by 1n in (i), we get:

2

1

2 2

15 1 3 1

2

1 15 10 5 3 3 5 7 2

2 2

nS n n

n n n n n

1n n nT S S

2 21 15 3 5 7 2

2 2

110 2 5 1 ........

2

n n n n

n n ii

Putting 20n in (ii), we get

20 5 20 1 99T

Hence, the 20th term is 99.

7. The sum of the first n term sofa an AP is 23 5

2 2

n n

. Find its nth term and the 25th term

Sol:


23 5

2 2n

n nS

2

1

3 1 5 1

2 2n

n nS

2

2

3 2 1 5 1

2 2

3 2

2

n n n

n n

thn term of the AP, na

1n nS S


______________________________________________________________________________

2 23 5 3 2

2 2

n n n n

6 2

2

3 2

n

n

Putting 25,n we get

25 3 25 1 75 1 76a

Hence, the nth term is 3 1n and 25th term is 76.

8. How many terms of the AP 21, 18, 15, … must be added to get the sum 0?

Sol:

Thee given AP is 21, 18, 15,……….


Let the required number of terms be n. Then,

0

2 21 1 3 0 2 12 2

42 3 3 02

n

n

S

n nn S a n d

nn

45 3 0

0 45 3 0

n n

n or n

0 15n or n

15n (Number of terms cannot be zero)

Hence, the required number of terms is 15.

9. How many terms of the AP 9, 17, 25, … must be taken so that their sum is 636?

Sol:

The given AP is 9, 17, 25,……



636nS

2 9 1 8 636 2 12 2

n

n nn S a n d

18 8 8 6362

nn

10 8 6362

nn


______________________________________________________________________________

2

2

4 5 636 0

4 48 53 636 0

4 12 53 12 0

12 4 53 0

12 0 4 53 0

5312

4

n n

n n n

n n n

n n

n or n

n or n

12n (Number of terms cannot negative)

Hence, the required number of terms is 12.

10. How many terms of the AP 63, 60, 57, 54, ….. must be taken so that their sum is 693?

Explain the double answer.

Sol:

The given AP is 63, 60, 57, 54,………..



693

2 63 1 3 693 2 12 2

126 3 3 6932

129 3 1386

n

n

S

n nn S a n d

nn

n n

2

2

3 129 1386 0

3 66 63 1386 0

n n

n n n

3 22 63 22 0

22 3 63 0

22 0 3 63 0

22 21

n n n

n n

n or n

n or n

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of

the AP is 0.

22 63 22 1 3 63 63 0a

Hence, the required number of terms is 21 or 22.

5 4 636n n


______________________________________________________________________________

11. How many terms of the AP 1 2

20,19 ,18 ,...3 3

must be taken so that their sum is 300? Explain

the double answer.

Sol:

The given AP is 20, 1 2

19 ,18 ,......3 3

Here, 20a and 1 58 58 60 2

19 20 203 3 3 3

d


300

22 20 1 300 2 1

2 3 2

2 240 300

2 3 3

122 2300

2 3

n

n

S

n nn S a n d

nn

nn

2

2

2

122 2 1800

2 122 1800 0

2 50 72 1800 0

2 25 72 25 0

25 2 72 0

25 0 2 72 0

25 36

n n

n n

n n n

n n n

n n

n or n

n or n

So, the sum of first 25 terms as well as that of first 36 terms is 300. This is because the sum

of all terms from 26th to 36th is 0.

12. Find the sum of all odd numbers between 0 and 50.

Sol:

All odd numbers between 0 and 50 are 1, 3, 5, 7,……… 49.

This is an AP in which 1, 3 1 2a d and 49.l

Let the number of terms be n.

Then, 49nT

1 49a n d

1 1 2 49

2 50

25

n

n

n


______________________________________________________________________________

Required sum 2

na l

25

1 49 25 25 6252

Hence, the required sum is 625.

13. Find the sum of all natural numbers between 200 and 400 which are divisible by 7.

Sol:

Natural numbers between 200 and 400 which are divisible by 7 are 203, 210 ,.... 399.

This is an AP with a = 203, d = 7 and l = 399.

Suppose there are n terms in the AP. Then,

399

203 1 7 399 1

7 196 399

7 399 196 203

29

n

n

a

n a a n d

n

n

n

Required sum 29

203 3992 2

n

nS a l

29602

2

8729


14. Find the sum of first forty positive integers divisible by 6.

Sol:

The positive integers divisible by 6 are 6, 12, 18,…..

This is an AP with 6a and 6.d

Also, 40n (Given)


n

nS a n d we get

40

402 6 40 1 6

2

20 12 234

20 246

4920

S



______________________________________________________________________________

15. Find the sum of first 15 multiples of 8.

Sol:

The first 15 multiples of 8 are 8, 16, 24, 32,…….

This is an AP in which 8, 16 8 8a d and 15.n

Thus, we have:

1

8 15 1 8

120

l a n d

Required sum 2

na l

15

8 120 15 64 9602


16. Find the sum of all multiples of 9 lying between 300 and 700.

Sol:

The multiples of 9 lying between 300 and 700 are 306, 315,………, 693.

This is an AP with 306, 9a d and 693.l

Suppose these are n terms in the AP. Then,

693

306 1 9 693 1

9 297 693

9 693 297 396

44

n

n

a

n a a n d

n

n

n

Required sum 44

306 6932 2

n

nS a l

22 999

21978


17. Find the sum of all three-digits natural numbers which are divisible by 13.

Sol:

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,.…… 938.

This is an AP in which a = 104, d = (117 – 104) = 13 and l = 938

Let the number of terms be n

Then 938nT


______________________________________________________________________________

1 988

104 1 13 988

13 897

69

a n d

n

n

n

Required sum 2

na l

69

104 988 69 546 376742


18. Find the sum of first 100 even number which are divisible by 5.

Sol:

The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...

This is an AP in which a = 10, d = (20 – l0) = 10 and n = 100

The sum of n terms of an AP is given by

2 12

1002 10 100 1 10 10, 10 100

2

50 20 990 50 1010 50500

n

nS a n d

a d and n

Hence, the sum of the first hundred even natural numbers which are divisible by 5 is

50500.

19. Find the sum of the following.

1 2 31 1 1 ...

n n n

up to n terms.

Sol:

On simplifying the given series, we get:

1 2 31 1 1 ...n

n n n

terms

1 2 3

1 1 1 ....... ........n

n termsn n n n

1 2 3........

nn

n n n n

Here, 1 2 3

......n

n n n n

is an AP whose first term is

1

nand the common difference


______________________________________________________________________________

is 2 1 1

.n n n

The sum of terms of an AP is given by

2 12

1 12 1

2

2 1 1

2 2

1 1

2 2

n

nS a n d

nn n

n n

n n n nn n

n n n

n nn

20. In an AP. It is given that 5 7 167S S and 10 235S , then find the AP, where nS denotes the

sum of its first n terms.

Sol:

Let a be the first term and d be the common difference of thee AP. Then,

5 7 167

5 72 4 2 6 167 2 1

2 2 2

5 10 7 21 167

12 31 167 ............ 1

n

S S

na d a d S a n d

a d a d

a d

Also,

10 235

102 9 235

2

5 2 9 235

2 9 47

S

a d

a d

a d

Multiplying both sides by 6, we get

12 54 282 ........... 2a d

Subtracting (1) from (2), we get

12 54 12 31 282 167

23 115

5

a d a d

d

d



______________________________________________________________________________

12 31 5 167

12 155 167

12 167 155 12

1

a

a

a

a

Hence, the AP is 1, 6, 11, 16,…….

21. In an AP, the first term is 2, the last term is 29 and the sum of all the terms is 155. Find the

common difference.

Sol:

Here, 2, 29 155na l and S

Let d be the common difference of the given AP and n be the total number of terms.

Then, 29nT

1 29

2 1 29 .......

a n d

n d i


1552

2 29 31 1552 2

10

n

nS a l

n n

n

Putting the value of n in (i), we get:

2 9 29

9 27

3

d

d

d

Thus, the common difference of the given AP is 3.

22. In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its

common difference.

Sol:

Suppose there are n terms in the AP.

Here, 4, 29a l and 150nS

150

4 29 1502 2

n

n

S

n nS a l

150 212

25n

Thus, the AP contains 12 terms.



______________________________________________________________________________

12 29

4 12 1 29 1

11 29 4 33

3

n

a

d a a n d

d

d


23. The first and the last terms of an AP are 17 and 350 respectively. If the common difference

is 9, how many terms are there and what is their sum?

Sol:


Here, a = 17, d = 9 and 350l

350

17 1 9 350 1

9 8 350

9 350 8 342

38

n

n

a

n a a n d

n

n

n

Thus, there are 38 terms in the AP.

38

2817 350

2 2

19 367

6973

n

nS S a l


24. The first and last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400,

find the common difference and the number of terms.

Sol:

Suppose there are n term in the AP.

Here, 5, 45a l and 400nS

400

5 45 4002 2

50 4002

400 216

50

n

n

S

n nS a l

n

n




______________________________________________________________________________

16 45

5 16 1 45 1

15 45 5 40

40 8

15 3

n

a

d a a n d

d

d

Hence, the common difference of the AP is 8

.3

25. In an AP, the first term is 22, nth terms is -11 and sum of first n terms is 66. Find the n and

hence find the4 common difference.

Sol:

Here, 22, 11na T and 66nS

Let d be the common difference of the given AP.

Then, 11nT

1 22 1 11

1 33 ........

a n dd n d

n d i


2 1 662

n

nS a n d [Substituting the value off 1n d from (i)]

2 22 33 11 662 2

12

n n

n

Putting the value of n in (i), we get:

11 33

3

d

d

Thus, 12n and 3d

26. The 12th term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its

first 10 terms.

Sol:


12 13

11 13 ....... 1 1n

a

a d a a n d

Also,

4 24S


______________________________________________________________________________

4

2 3 24 2 12 2

n

na d S a n d

2 3 12 ........ 2a d

Solving (1) and (2), we get

2 13 11 3 12

26 22 3 12

19 12 26 38

2

d d

d d

d

d


11 2 13

13 22 9

a

a

Sum of its first 10 terms, 10S

102 9 10 1 2

2

5 18 18

5 0

0


27. The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1:5,

find the AP.

Sol:

Let a be the first term and d be the common difference of the AP.

7 182

72 6 182 2 1

2 2

3 26 .......... 1

n

S

na d S a n d

a d

Also,

4 17: 1:5a a (Given)

3 11

16 5

5 15 16

4 ......... 2

n

a da a n d

a d

a d a d

d a



______________________________________________________________________________

3 4 26

13 26

2

a a

a

a

Putting 2a in (2), we get

4 2 8d

Hence, the required AP is 2, 10, 18, 26,………

28. The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first

term and common difference of the AP.

Sol:

Here, 4, 7 81a d and l

Let the nth term be 81.

Then 81nT

1 4 1 7 81

1 7 77

1 11

12

a n d n

n

n

n



12

2

124 81 6 85 510

2

n

nS a l

S

Thus, the required sum is 510.

29. The sum of the first 7 terms of an AP is 49 and the sum of its first 17 term is 289. Find the

sum of its first n terms.

Sol:

Let a be the first term and d be the common difference of the given AP.

Then, we have:

7

17

2 12

72 6 7 3

2

172 16 17 8

2

n

nS a n d

S a d a d

S a d a d

However, 7 49S and 17 289S

Now, 7 3 49a d


______________________________________________________________________________

3 7 ........a d i

Also, 17 8 289a d

8 17 .........a d ii

Subtracting (i) from (ii), we get:

5 10

2

d

d

Putting 2d in (i), we get

6 7

1

a

a

Thus, 1a and 2d

Sum of n terms of AP = 22 1 1 2 1 12

nn n n n

30. Two Aps have the same common difference. If the fist terms of these Aps be 3 and 8

respectively. Find the difference between the sums of their first 50 terms.

Sol:

Let 1a and 2a be the first terms of the two .APs

Here, 1 28 3a and a

Suppose d be the common difference of the two Aps

Let 50S and 50 1 2

50 50' 2 50 1 2 50 1

2 2S a d a d

25 2 8 49 25 2 3 49

25 16 6

250

d d

Hence, the required difference between the two sums is 250.

31. The sum first 10 terms of an AP is -150 and the sum of its next 10 terms is -550 . Find the

AP.

Sol:


10 150S (Given)

102 9 150 2 1

2 2

5 2 9 150

2 9 30 ........ 1

n

na d S a n d

a d

a d

It is given that the sum of its next 10 terms is 550.


______________________________________________________________________________

Now,

20S Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms =

150 550 700

20 700

202 19 700

2

10 2 19 700

2 19 70 ......... 2

S

a d

a d

a d


2 19 2 9 70 30

10 40

4

a d a d

d

d


2 9 4 30

2 30 36 6

3

a

a

a

Hence, the required AP is 3, 1, 5, 9,.............

32. The 13th terms of an AP is 4 times its 3rd term. If its 5th term is 16, Find the sum of its first

10 terms.

Sol:


13 34a a (Given)

12 4 2 1

12 4 8

3 4 ........... 1

na d a d a a n d

a d a d

a d

Also,

5 16a (Given)

4 16 .............(2)a d


3 16

4 16

4

a a

a

a


4 3 4 12

3

d

d


______________________________________________________________________________

Using the formula, 4 2 1 ,2

nS a n d we get

10

102 4 10 1 3

2

5 8 27

5 35

175

S


33. The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first

15 terms.

Sol:


16 35a a (Given)

15 5 2 1na d a d a a n d

15 5 10

4 5

a d a d

a d

Also,

10 41a (Given)

9 41 .......... 2a d


49 41

5

5 3641

5

4141

5

5

aa

a a

a

a


5 4 5 20

4

d

d


n

nS a n d we get


______________________________________________________________________________

15

152 5 15 1 4

2

1510 56

2

1566

2

495

S


34. An AP 5, 12, 19, .... has 50 term. Find its last term. Hence, find the sum of its last 15 terms.

Sol:

The given AP is 5,12,19,.......

Here, 5, 12 5 7a d and 50.n

Since there are 50 terms in the AP, so the last term of the AP is 50 .a

50 5 50 1 7 1nl a a a n d

5 343

348

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP

50 35

50 352 5 50 1 7 2 5 35 1 7

2 2

2 12

50 3510 343 10 238

2 2

50 35353 248

2 2

n

S S

nS a n d

17650 8680

2

8970

2

4485

Hence, the require sum is 4485.

35. An AP 8, 10, 12, … has 60 terms. Find its last term. Hence, find the sum of its last 10

terms.

Sol:



______________________________________________________________________________

Here, 8, 10 8 2a d and 60n


60 8 60 1 2 1

8 118

126

nl a a a n d


Now,

Sum of the last 10 terms of the AP

60 50

60 502 8 60 1 2 2 8 50 1 2

2 2

S S

2 12

30 16 118 25 16 98

30 134 25 114

4020 2850

1170

n

nS a n d


36. The sum of the 4th and 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44.

Find the sum of its first 10 terms.

Sol:

Let a be the first and d be the common difference of the AP.

4 8 24a a (Given)

3 7 24 1

2 10 24

5 12 ...... 1

na d a d a a n d

a d

a d

Also,

6 10 44a a (Given)

5 9 44 1

2 14 44

7 22 ...... 2

na d a d a a n d

a d

a d


7 5 22 12a d a d

2 10d

5d


______________________________________________________________________________


5 5 12

12 25 13

a

a


n

nS a n d we get

10

102 13 10 1 5

2

5 26 45

5 19

95

S


37. The sum of fist m terms of an AP is ( 24m m ). If its nth term is 107, find the value of n. Also,

Find the 21st term of this AP.

Sol:

Let mS denotes the sum of the first m terms of the AP. Then,

2

2

1

2

2

4

4 1 1

4 2 1 1

4 9 5

m

m

S m m

S m m

m m m

m m

Suppose ma denote the thm term of the AP.

1

2 24 4 9 5

8 5 ....... 1

m m ma S S

m m m m

m

Now,

107na (Given)

8 5 107n [From (1)]

8 107 5 112

14

n

n

Thus, the value of n is 14.

Putting 21m in (1), we get

21 8 21 5 168 5 163a

Hence, the 21st term of the AP is 163.


______________________________________________________________________________

38. The sum of first q terms of an AP is 263 3q q . If its pth term is -60, find the value of p. Also,

find the 11th term of its AP.

Sol:

Let qS denote the sum of the first q terms of the AP. Then,

2

2

1

2

2

63 3

63 1 3 1

63 63 3 2 1

3 69 66

q

q

S q q

S q q

q q q

q q

Suppose qa denote the thq term of the AP.

1

2 263 3 3 69 66

6 66 ........ 1

q q qa S S

q q q q

q

Now,

60pa (Given)

6 66 60p [From (1)]

6 60 66 126

21

p

p

Thus, the value of p is 21.

Putting 11q in (1), we get

11 6 11 66 66 66 0a

Hence, the 11th term of the AP is 0.

39. Find the number of terms of the AP -12, -9, -6, .., 21. If 1 is added to each term of this AP

then the sum of all terms of the AP thus obtained.

Sol:

The given AP is 12, 9, 6,.....,21.

Here, 12, 9 12 9 12 3 2a d and l l


21

12 1 3 21 1

3 15 21

3 21 15 36

12

n

n

l a

n a a n d

n

n

n



______________________________________________________________________________

If 1 is added to each term of the AP, then the new AP so obtained is 11, 8, 5,........,22.

Here, first term, 11;A last term, 22 12 L and n

Sum of the terms of this AP

12

11 222 2

6 11

66

n

nS a l


40. Sum of the first 14 terms of and AP is 1505 and its first term is 10. Find its 25th term.

Sol:


Here, 10a and 14n

Now,

14 1505S (Given)

142 10 14 1 1505 2 1

2 2

7 20 13 1505

20 13 215

13 215 20 195

15

n

nd S a n d

d

d

d

d

25th term of the AP, 25a

10 25 1 15 1

10 360

370

na a n d

Hence, the required term is 370.

41. Find the sum of fist 51 terms of an AP whose second and third terms are 14 and 18

respectively.

Sol:


3 2 18 14 4d a a

Now,

2 14a (Given)

14 1

4 14

14 4 10

na d a a n d

a

a


______________________________________________________________________________


n

nS a n d we get

51

512 10 51 1 4

2

5120 200

2

51220

2

5610

S


42. In a school, students decided to plant trees in and around the school to reduce air pollution.

It was decided that the number of trees that each section of each class will plant will be

double of the class in which they are studying. If there are 1 to 12 classes in the school and

each class has two section, find how many trees were planted by student. Which value is

shown in the question?

Sol:

Number of trees planted by the students of each section of class 1 = 2

There are two sections of class 1.

Number of trees planted by the students of class 1 2 2 4

Number of trees planted by the students of each section of class 2 = 4

There are two sections of class 2.


Similarly,


So, the number of trees planted by the students of different classes are 4, 8, 12, ....

Total number of trees planted by the students 4 8 12 ...... up to 12 terms

This series is an arithmetic series.

Here, 4, 8 4 4 12a d and n


n

nS a n d we get

12

122 4 12 1 4

2

6 8 44

6 52

312

S

Hence, the total number of trees planted by the students is 312.

The values shown in the question are social responsibility and awareness for conserving

nature.


______________________________________________________________________________

43. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,

and the other potatoes are placed 3m apart in a straight line. There are 10 potatoes in the

line. A competitor starts from the bucket, picks up the nearest potato, runs back with it,

drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in,

and he continues in the same way until all the potatoes are in the bucket. What is the total

distance the competitor has to run?

Sol:

Distance covered by the competitor to pick and drop the first potato 2 5 10m m

Distance covered by the competitor to pick and drop the second potato

2 5 3 2 8 16m m m

Distance covered by the competitor to pick and drop the third potato

2 5 3 3 2 11 22m m m and so on.

Total distance covered by the competitor 10 16 22 ......m m m up to 10 terms

This is an arithmetic series.

Here, 10, 16 10 6a d and 10n


n

nS a n d we get

10

102 10 10 1 6

2

5 20 54

5 74

370

S

Hence, the total distance the competitor has to run is 370 m.

44. There are 25 trees at equal distance of 5 m in a line with a water tank, the distance of the

water tank from the nearest tree being 10 m. A gardener waters all the trees separately,

starting from the water tank and returning back to the water tank after watering each tree to

get water for the next. Find the total distance covered by the gardener in order to water all

the trees.


______________________________________________________________________________

Sol:

Distance covered by the gardener to water the first tree and return to the water tank

10 10 20m m m

Distance covered by the gardener to water the second tree and return to the water tank

15 15 30m m m

Distance covered by the gardener to water the third tree and return to the water tank

20 20 40m m m and so on.

Total distance covered by the gardener to water all the trees 20 30 40 ......m m m up

to 25 terms

This series is an arithmetic series.

Here, 20, 30 20 10 25a d and n


n

nS a n d we get

25

252 20 25 1 10

2

2540 240

2

25280

2

3500

S

Hence, the total distance covered by the gardener to water all the trees 3500 m.

45. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their

overall academic performance. If each prize is ₹20 less than its preceding prize, find the

value of each prize.

Sol:

Let the value of the first prize be a.

Since the value of each prize is 20 less than its preceding prize, so the values of the prizes

are in AP with common difference – ₹20.


______________________________________________________________________________

d ₹

402 40 1 36000

2

20 2 39 36000

2 39 1800 ........ 2

a d

a d

a d

Number of cash prizes to be given to the students, n = 7

Total sum of the prizes, 7 S ₹700


n

nS a n d we get

7

72 7 1 20 700

2

72 120 700

2

7 420 700

7 700 420 1120

160

S a

a

a

a

a

Thus, the value of the first prize is ₹160.

Hence, the value of each prize is ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.

46. A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more

than he did in the preceding month. How much did he save in the first month?

Sol:

Let the money saved by the man in the first month be ₹a

It is given that in each month after the first, he saved ₹100 more than he did in the

preceding month. So, the money saved by the man every month is in AP with common

difference ₹100.

d ₹100

Number of months, n = 10

Sum of money saved in 10 months, 10S ₹ 33,000


n

nS a n d we get

10

102 10 1 100 33000

2

5 2 900 33000

2 900 6600

2 6600 900 5700

2850

S a

a

a

a

a

Hence, the money saved by the man in the first month is ₹2,850.


______________________________________________________________________________

47. A man arranges to pay off debt of ₹36000 by 40 monthly instalments which form an

arithmetic series. When 30 of the installments are paid, he dies leaving on-third of the debt

unpaid. Find the value of the first instalment.

Sol:

Let the value of the first installment be ₹a.

Since the monthly installments form an arithmetic series, so let us suppose the man

increases the value of each installment by ₹d every month.

Common difference of the arithmetic series = ₹d

Amount paid in 30 installments = ₹36,0001

3 ₹ 36,000 ₹ 36,000 ₹ 12,000 ₹ 24,000

Let nS denote the total amount of money paid in the n installments. Then,

30 24,000

302 30 1 24000 2 1

2 2

15 2 29 24000

2 29 1600 ......... 1

n

S

na d S a n d

a d

a d

Also,

40S ₹36,000

402 40 1 36000

2

20 2 39 36000

2 39 1800 ........ 2

a d

a d

a d


2 39 2 29 1800 1600

10 200

20

a d a d

d

d


2 29 20 1600

2 580 1600

2 1600 580 1020

510

a

a

a

a

Thus, the value of the first installment is ₹510.

48. A contract on construction job specifies a penalty for delay of completion beyond a certain

date as follows: ₹200 for the first day, ₹ 250 for the second day, ₹300 for the third day, etc.

the penalty for each succeeding day being ₹50 more than for the preceding day. How much

money the contractor has to pay as penalty, if he has delayed the work by 30 days?


______________________________________________________________________________

Sol:

It is given that the penalty for each succeeding day is 50 more than for the preceding day,

so the amount of penalties are in AP with common difference ₹50

Number of days in the delay of the work = 30

The amount of penalties are ₹200, ₹250, ₹300,... up to 30 terms.

Total amount of money paid by the contractor as penalty,

30S ₹ 200 ₹ 250 ₹ 300 ....... up to 30 terms

Here, a ₹ 200,d ₹ 50 30and n


n

nS a n d we get

30

302 200 30 1 50

2

15 400 1450

S

15 1850

27750

Hence, the contractor has to pay ₹27,750 as penalty

Exercise - Multiple Choice Questions

1. The common difference of the AP 1 1 1 2

, , ,.......2

p p

p p

is

(a) p (b) –p (c) -1 (d) 1

Answer: (c) -1

Sol:

The given AP is 1 1 1 2

, , ,.......2

p p

p p

Common difference, 1 1 1 1

1p p p

dp p p p

2. The common difference of the AP 1 1 3 1 6

, , ,.....3 3 3

b b is

(a) 1

3(b)

1

3

(c) b (d) –b

Answer: (d) –b

Sol:

The given AP is 1 1 3 1 6

, , ,.....3 3 3

b b


______________________________________________________________________________

Common difference, 1 3 1 1 3 1 3

3 3 3 3

b b bd b

3. The next term of the AP 7, 28, 63,... is

(a) 70 (b) 84 (c) 98 (d) 112

Answer: (d) 112

Sol:

The given terms of the AP can be written as 7, 4 7, 9 7,........ . . 7,2 7,3 7,......i e

Next term 4 7 16 7 112

4. If 1 2 3,4, , , 28x x x are in AP then 3, ?x

(a) 19 (b) 23 (c) 22 (d) cannot be determined

Answer: (c) 22

Sol:

Here, 4, 28 5a l and n

Then, 5 28T

1 28

4 5 1 28

4 24

6

a n d

d

d

d

Hence, 3 28 6 22x

5. If the nth term of an AP is (2n + 1) then the sum of its first three terms is

(a) 6n+3 (b) 15 (c) 12 (d) 21

Answer: (b) 15

Sol:

nth term of the AP, 2 1na n (Given)

First term, 1 2 1 1 2 1 3a

Second term, 2 2 2 1 4 1 5a

Third term, 3 2 3 1 6 1 7a

Sum of the first three terms 1 2 3 3 5 7 15a a a


______________________________________________________________________________

6. The sum of first n terms of an AP is 23 6 .n n The common difference of the AP is

(a) 6 (b) 9 (c) 15 (d) -3

Sol:


2

2

1

2

2

3 6

3 1 6 1

3 2 1 6 1

3 3

n

n

S n n

S n n

n n n

n

So, thn term of the AP, 1n n na S S

2 23 6 3 3

6 3

n n n

n


1

6 3 6 1 3

6 3 6 1 3

6

n nd a a

n n

n n

Thus, the common difference of the AP is 6.

7. The sum of first n terms of an AP is 25 .n n The nth term of the AP is

(a) ( 5 - 2n) (b) ( 6 – 2n) (c) (2n – 5) (d) (2n – 6)

Answer: (b) ( 6 – 2n)

Sol:


2

2

1

5

5 1 1

n

n

S n n

S n n

2

2

5 5 2 1

7 6

n n n

n n

thn term of the AP, 1n n na S S

2 25 7 6

6 2

n n n n

n

Thus, the nth term of the AP is 6 2 .n


______________________________________________________________________________

8. The sum of the first n terms of an AP is 24 2 .n n The nth term of this AP is

(a) (6n - 2) (b) (7n – 3) (c) (8n – 2) (d) (8n + 2)

Answer: (c) (8n – 2)

Sol:


2

2

1

2

2

4 2

4 1 2 1

4 2 1 2 1

4 6 2

n

n

S n n

S n n

n n n

n n

thn term of the AP, 1n n na S S

2 24 2 4 6 2

8 2

n n n n

n

Thus, the thn term of thee AP is 8 2n

9. The 7th term of an AP is -1 and its 16th term is 17. The nth term of the AP is

(a) (3n + 8) (b) (4n – 7) (c) (15 – 2n) (d) (2n – 15)

Answer: (d) (2n – 15)

Sol:


nth term of the AP, 1na a n d

Now,

7 1a (Given)

6 1 ......... 1a d

Also,

16 17a (Given)

15 17 ........ 2a d


15 6 17 1

9 18

2

a d a d

d

d

Putting d = 2 in (1), we get

6 2 1

1 12 13

a

a

nth term of the AP, 13 1 2 2 15na n n


______________________________________________________________________________

10. The 5th term of an AP is -3 and its common difference is -4. The sum of the first 10 terms is

(a) 50 (b) -50 (c) 30 (d) -30

Answer: (b) -50

Sol:

Let a be the first term of the AP.

Here. 4d

5 3a (Given)

5 1 4 3 1

16 3

16 3 13

na a a n d

a

a


n

nS a n d we get

10

102 13 10 1 4

2

5 26 36

5 10

50

S

Thus, the sum of its first 10 terms is 50.

11. The 5th term of an AP is 20 and the sum of its 7th and 11th terms is 64. The common

difference of the AP is

(a) 4 (b) 5 (c) 3 (d) 2

Answer: (c) 3

Sol:


5 20

5 1 20 1

4 20 ........... 1

n

a

a d a a n d

a d

Now,

7 11 64a a (Given)

6 10 64

2 16 64

8 32 .......... 2

a d a d

a d

a d


20 4 8 32d d

4 32 20 12d

3d


______________________________________________________________________________


12. The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten

terms is

(a) 150 (b) 175 (c) 160 (d) 135

Answer: (b) 175

Sol:


13 34a a (Given)

12 4 2 1

12 4 8

3 4 ............ 1

na d a d a a n d

a d a d

a d

Also

5 16a (Given)

4 16 ........ 2a d


3 16

4 16

4

a a

a

a

Putting a = 4 in (1), we get

4 3 4 12

3

d

d


n

nS a n d we get

10

102 4 10 1 3

2

5 8 27

5 35

175

S

Thus, the sum of its first 10 terms is 175.

13. An AP 5, 12, 9, …. has 50 terms. Its last term is

(a) 343 (b) 353 (c) 348 (d) 362

Answer: (c) 348

Sol:


Here, 5, 12 5 7a d and 50n


______________________________________________________________________________


50 5 50 1 7 1

5 343

348

na a a n d


14. The sum of the first 20 odd natural numbers is

(a) 100 (b) 210 (c) 400 (d) 420

Answer: (c) 400

Sol:

The first 20 odd natural numbers are 1, 3, 5,.., 39.

These numbers are in AP.

Here. a = 1, l = 39 and n = 20

Sum of first 20 odd natural numbers

20

1 392 2

10 40

400

n

nS a l

15. The sum of first 40 positive integers divisible by 6 is

(a) 2460 (b) 3640 (c) 4920 (d) 4860

Answer: (c) 4920

Sol:

The positive integers divisible by 6 are 6, 12, 18,....

This is an AP with a = 6 and d = 6.

Also, n = 40 (Given)


n

nS a n d we get

40

402 6 40 1 6

2

20 12 234

20 246

4920

S

Thus, the required sum is 4920

16. How many two digit numbers are divisible by 3?

(a) 25 (b) 30 (c) 32 (d) 36

Answer: (b) 30

Sol:


______________________________________________________________________________

The two-digit numbers divisible by 3 are 12, 15, 18,…… 99.


Here, a = 12 and d = 15 12 3

Let this AP contains n terms. Then.

99

12 1 3 99 1

3 9 99

3 99 9 90

30

n

n

a

n a a n d

n

n

n

Thus, there are 30 two-digit numbers divisible by 3.

17. How many three-digit number are divisible by 9?

(a) 86 (b) 90 (c) 96 (d) 100

Answer: (d) 100

Sol:

The three-digit numbers divisible by 9 are 108, 117, 126 ,…. 999.

Clearly, these numbers are in AP.

Here, 108 117 108 9 a and d


999

108 1 9 999 1

9 99 999

9 999 99 900

100

n

n

a

n a a n d

n

n

n

Thus, there are 100 three-digit numbers divisible by 9.

18. What is the common difference of an AP in which 18 14 32a a ?

(a) 8 (b) -8 (c) 4 (d) -4

Answer: (a) 8

Sol:

Let a be the first term and d be the common difference of the AP. Then.

18 14 32

17 13 32 1

4 32

8

n

a a

a d a d a a n d

d

d



______________________________________________________________________________

19. If na denotes the nth term of the AP 3, 8, 13, 18, … then what is the value of 30 20a a ?

(a) 40 (b) 36 (c) 50 (d) 56

Answer: (c) 50

Sol:

The given AP is 3, 8,13,18....

Here, a = 3 and d = 8 – 3 = 5

30 20

3 30 1 5 3 20 1 5 1n

a a

a a n d

148 98

50

Thus, the required value is 50.

20. Which term of the AP 72, 63, 54, …. is 0 ?

(a) 8th (b) 9th (c) 10th (d) 11th

Answer: (b) 9th

Sol:

The given AP is 72, 63, 54,.....

Here, a = 72 and d = 63 – 72 = –9

Suppose nth term of the given AP is 0. Then.

0

72 1 9 0 1

9 81 0

819

9

n

n

a

n a a n d

n

n

Thus, the 9th term of the given AP is 0.

21. Which term of the AP 25, 20, 15, … is the first negative term?

(a) 10th (b) 9th (c) 8th (d) 7th

Answer: (d) 7th

Sol:


Here, a = 25 and 20 25 5d


0

25 1 5 0 1

30 5 0

n

n

a

n a a n d

n

5 30n


______________________________________________________________________________

306

5n

7n

Thus, the 7th term is the first negative term of the given AP.

22. Which term of the AP 21, 42, 63, 84, … is 210?

(a) 9th (b) 10th (c) 11th (d) 12th

Answer: (b) 10th

Sol:

Here, 21a and 42 21 21d

Let 210 be the nth term of the given AP.

Then, 210nT

1 210

21 1 21 210

21 210

10

a n d

n

n

n

Hence, 210 is the 10th term of the AP.

23. What is 20th term form the end of the AP 3, 8, 13, …., 253?

(a) 163 (b) 158 (c) 153 (d) 148

Answer: (b) 158

Sol:

The given AP is 3, 8, 13,..., 253.

Let us re-write the given AP in reverse order i.e. 253, 248, .., 13, 8, 3.

Now, the 20th term from the end of the given AP is equal to the 20th term from beginning

of the AP 253, 248,., 13, 8, 3.

Consider the AP 253, 248,….. 13, 8, 3.

Here, a = 253 and 248 253 5d

20th term of this AP

253 20 1 5

253 95

158

Thus, the 20th term from the end of the given AP is 158.

24. (5 + 13 + 21 + … + 181) = ?

(a) 2476 (b) 2337 (c) 2219 (d) 2139

Answer: (d) 2139

Sol:


______________________________________________________________________________

Here, 5, 13 5 8a d and 181l

Let the number of terms be n.

Then, 181nT

1 181

5 1 8 181

8 184

23

a n d

n

n

n

Required sum 2

na l

23

5 181 23 93 21392


25. The sum of first 16 terms of the AP 10, 6, 2, …. is

(a) 320 (b) -320 (c) -352 (d) -400

Answer: (b) -320

Sol:

Here, 10, 6 10 4 16a d and n


n

nS a n d we get

16

162 10 16 1 4

2S

10, 4 16a d and n

8 20 60 8 40 320

Hence, the sum of the first 16 terms of the given AP is -320.

26. How many terms of the AP 3, 7, 11, 15, … will make the sum 406?

(a) 10 (b) 12 (c) 14 (d) 20

Answer: (c) 14

Sol:

Here, 3a and 7 3 4d

Let the sum of n terms be 406.

Then, we have:

2 1 4062

2 3 1 4 4062

n

nS a n d

nn


______________________________________________________________________________

2

2

2

3 2 2 406

2 28 29 406

2 406 0

2 28 29 406 0

2 14 29 14 0

2 29 14 0

n n

n n n

n n

n n n

n n n

n n

14n ( n can’t be a fraction)

Hence, 14 terms will make the sum 406.

27. The 2nd term of an AP is 13 and 5th term is 25. What is its 17 term?

(a) 69 (b) 73 (c) 77 (d) 81

Answer: (b) 73

Sol:

2

5

13 ........

4 25 ........

T a d i

T a d ii


3 12

4

d

d

On putting the value of d in (i), we get:

4 13

9

a

a

Now, 17 16 9 16 4 73T a d

Hence, the 17th term is 73.

28. The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is

(a) 3 (b) 2 (c) -3 (d) -2

Answer: (a) 3

Sol:

10

17

9

16

T a d

T a d

Also, 1016 21a d T

16 21 9

7 21

3

a d d

d

d



______________________________________________________________________________

29. The 8th term of an AP is 17 and its 14th term is 29. The common difference of the AP is

(a) 3 (b) 2 (c) 5 (d) -2

Answer: (b) 2

Sol:

8

14

7 17 ......

13 29 ......

T a d i

T a d ii


6 12

12

d

d

Hence, the common difference is 2.

30. The 7th term of an AP is 4 and its common difference is -4. What is its first term?

(a) 16 (b) 20 (c) 24 (d) 28

Answer: (d) 28

Sol:

7 6

6 4 4

4 24 28

T a d

a

a

Hence, the first term us 28.

class x chapter 11 arithmetic progression maths

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